Supermartingale vanishing at some stopping time Let $\left\{X_t\right\}_{t\in[0, T]}$ be a continuous and non-negative supermartingale. We define the stopping time 
$$\tau_0:=\inf\{t\in[0,T]:X(t)=0\}\wedge T$$
and immediately obtain by continuity of $X$ that $\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)=0$ a.s.
I want to prove that $X(t)=0$ for almost every $\omega\in\{\tau_0<T\}$ where $t\in[\tau_0(\omega),T]$.
What I established so far (using optional stopping) is the following:
$$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)\mid F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$
Hence, as $X$ is non-negative, the desired result follows for the terminal time $t=T$. Can someone please help me to prove the whole result for all the other relevant times? Thanks a lot in advance!
 A: You can generalize your derivation from:
$$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$
to:
$$E[\mathbf{1}_{\{\tau_0=s\}}X(t)]=E[\mathbf{1}_{\{\tau_0=s\}}E[X(t)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0=s\}}X(\tau_0)]=0$$
where $t > s$.
As you mentioned, using the fact that $X(t)$ is non-negative, you can imply state from there that:
$$E[\mathbf{1}_{\{\tau_0=s\}}X(t)] = \int_0^\infty x p(x_t = x \wedge \tau_0=s) dx \leq 0$$
This can only hold if $p(x_t = x \wedge \tau_0 = s) = 0$ for all $x > 0$. Therefore, for any path that has a stopping time $s$, $p(x_t > x) = 0$ for all $x > 0$. 

For future readers: How to finish the proof - For a fixed $s$ I have shown that $P(X(t) = 0|\tau_0 = s) = 1$ for all $t > s$.  Now take a countable dense subset $t_i$ of $(s, T]$.  Then $P(X(t_i) = 0 \,\,\,\ \forall t_i | \tau_0 = s) = 1$.  A continuous path that is zero at a countable dense set of points must be zero everywhere.
Now to finish:
$$P(X_t \neq 0 \,\,\, \textrm{for some} \,\,t > \tau_0 \wedge \tau_0 < T) = \int_0^T p(X_t \neq 0 \,\,\, \textrm{for some} \,\,t > s \wedge \tau_0 = s) ds = 0$$
A: You just need to look at more stopping times. For any $t\geq 0$, you have
$${E}[\mathbf{1}_{\{\tau_0<T\}}X(T \wedge (\tau_0+t))]
={E}[\mathbf{1}_{\{\tau_0<T\}}E[X(T\wedge (\tau_0+t))|F_{\tau_0}]]
\le {E}[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0.$$
This gives
$$P\biggl(X(T\wedge(\tau_0+t))=0\mbox{ for all }t\in\mathbb{Q}\cap[0,T]; \tau_0<T\biggr)
=P(\tau_0<T),$$
and sample path continuity will give you the desired conclusion.
A: Let $\epsilon > 0$ and consider the stopping time
$$\tau_\epsilon = \inf\{t \in [\tau_0, T] : X_t \ge \epsilon\} \wedge T.$$
Since $\tau_0 \le \tau_\epsilon$ almost surely, optional stopping gives $E[X_{\tau_\epsilon}] \le E[X_{\tau_0}]$.  Now let consider the events $\{\tau_0 < T\}$ and $\{\tau_0 = T\}$ and write
$$E[X_{\tau_\epsilon} ; \tau_0 < T] + E[X_{\tau_\epsilon} ; \tau_0 =T] \le E[X_{\tau_0} ; \tau_0 < T] + E[X_{\tau_0} ; \tau_0 =T].$$
Now $X_{\tau_0} = 0$ on the event $\{\tau_0 < T\}$.  On the event $\{\tau_0 = T\}$ we have $X_{\tau_0} = X_T$, and moreover, since $\tau_\epsilon \ge \tau_0$, we also have $\tau_\epsilon = T$ and $X_{\tau_\epsilon} = X_T$.  So our inequality becomes
$$E[X_{\tau_\epsilon} ; \tau_0 < T] + E[X_{T} ; \tau_0 =T] \le  0 + E[X_{T} ; \tau_0 =T]$$
showing that $E[X_{\tau_\epsilon} ; \tau_0 < T] \le 0$.  Since $X_{\tau_\epsilon}$ is nonnegative we have $X_{\tau_\epsilon} = 0$ almost everywhere on $\{\tau_0 < T\}$.  On the other hand, $\{\tau_\epsilon < T\} \subset \{\tau_0 < T\}$, and by continuity $X_{\tau_\epsilon} = \epsilon$ almost everywhere on $\{\tau_\epsilon < T\}$.  So we conclude $P(\tau_\epsilon < T) = 0$, i.e. almost surely on $\{\tau_0 < T\}$, there does not exist $t$ in $[\tau_0, T)$ with $X_t \ge \epsilon$.  Letting $\epsilon \downarrow 0$ along a sequence, we get that almost surely on $\{\tau_0 < T\}$, there does not exist $t \in [t_0, T)$ with $X_t > 0$.  This is the desired conclusion.
A: Here are two approaches, one more elementary than the other.


*

*Modify your argument to show that $X(r) = 0$ a.s on the event
$\{\tau_0< r\}$ for each $r\in (0,T]$. Deduce from this that $X(r)=0$ for every rational in $(\tau_0,T)$, a.s. on the event $\{\tau_0<T\}$. Now invoke the continuity of $X$.

*Let $B=\{\omega: \tau_0(\omega)<T, X(t,\omega)>0$ for some $t\in(\tau_0,T)\}$, and suppose that $P[B]>0$. Then by Meyer's Optional Section Theorem, there is a stopping time $S\ge\tau_0$ such that $P[S<\infty]>0$ and $X(S(\omega),\omega)>0$ for each $\omega\in\{S<\infty\}$. By optional stopping at time $S$, $E[X(S)]=0$ because $S\ge\tau_0$. On the other hand
$$
E[X(S)] \ge E[X(S); S<\infty] >0.
$$
This contradiction shows that $P[B]=0$.
